Question: Let $f$ be a differentiable function with $f(-3)=0$ and $f'(-3)=7$. What is the value of the approximation of $f(-3.2)$ using the function's local linear approximation at $x=-3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1.4$ (Choice B) B $-1.3$ (Choice C) C $-1.2$ (Choice D) D $-1.1$
Solution: The local linear approximation of $f$ at $x=-3$ is achieved using the equation of the line tangent to $f$ at $x=-3$. Let $L(x)$ represent this equation. We can find $L(x)$ using the general formula for the tangent to the graph of function $u$ at $x=a$ : $y=u'(a)(x-a)+u(a)$ [Is there a way to find this formula without memorizing?] In our case, $L(x)=f'(-3)(x+3)+f(-3)$. Plugging $f(-3)=0$ and $f'(-3)=7$, we obtain $L(x)=7(x+3)+0=7(x+3)$. To approximate $f(-3.2)$, all we need is to plug $x=-3.2$ into $L(x)$. $\begin{aligned} L(-3.2)&=7(-3.2+3) \\\\ &=7(-0.2) \\\\ &=-1.4 \end{aligned}$ In conclusion, the approximation of $f(-3.2)$ using the function's local linear approximation at $x=-3$ is $-1.4$.